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Which parts of an airplane are made of buoyant material?
Which parts of an airplane (to make this question more specific: accessible in the main fuselage where the passengers are aboard) are made of buoyant material.
Imagine a plane is about to crash into the sea, you might wanna hang on to something that could keep you afloat, so here’s a few things that I think are buoyant(based on the floating debris seen after a crash):

The life vest/jackets which we can inflate
Seats
The doors(the ones through which we enter or any other door)

Which things would someone want to hold onto, to increase chances of survival?

Passengers are reminded of #1 & 2 above on every safety brief.
Beyond that, you aren't going to get up out of your seat before the crash and attempt to latch on to some buoyant part of the structure, so it isn't an actionable question in the context it is asked.  (Which is a polite way of saying who cares...)
Stay in your seat until the violent motion stops, then follow the instructions of the surviving crew to evacuate the aircraft in an orderly manner.
Allow women and children to go first if able.  Whatever might be floating after the crash is over is buoyant, and may be used.

I know that the odds of something like this to occur are quite low, but let’s say that since we are aviation geeks and we know when an aircraft would have a massive hit killing all passengers on impact, we could at least try something to increase our chances of survival, by running to the end of the aircraft(for reducing the force impact) and latch yourself to something buoyant to prevent drowning while unconscious, why not try doing something like this, I mean what do we have to lose

@Valay_17 You have your life to lose. Even the most gentle ditching is likely to be fairly violent. You're much more likely to survive the initial impact, and have fewer major injuries, if you're seated with your seat belt fastened. Given the typical speed of an aircraft, "running backward" isn't going to make hardly any practical difference in impact force, and will do absolutely nothing for sideways and vertical forces.

Or, if you survive, you are brought up on charges for failure to obey the crew and endangering other passengers for your selfish and desperate actions.

@HiddenWindshield Running backwards as in go the end of the aircraft(and brace for impact) to reduce the head on impact(which might be enough to not knock someone unconscious). Also gentle ditching isn’t the case, I am assuming we have lost an engine/control and it is going to be quite rough where definitely the odds of survival are at the lower spectrum but maybe something that we do might increase our chances.

@Valay_17, so if everyone runs to the back at the same time does CG shift and the airplane stall?  Not everyone will fit, so you start fighting?  How do you know whether it will be survivable or not?  (Miracle on the Hudson)  Personally I would rather stay in my seat and die with dignity than scramble as you propose.  And if you were to try this you risk getting throat punched by a burly passenger in assistance of the crew.  I would, and I'm not burly...

@MichaelHall Not everyone would do that, and those who do would help each other, I mean survival is important so. Again this is something that has the least probability to happen, also it is gonna happen so fast people won’t have time to react to thus but rather focus on survival.

Survival of EVERYONE is important.  So, stay in your seat!!!

As an self-proclaimed aviation geek, @Valay_17, you should know that being tightly strapped into your seat gives you the best chance of surviving the initial impact. If you survive that, then you find anything floating and grab hold of that. If you're bobbing in the sea, you're really not going to care one bit what floating thing you're holding on to. You're just hoping the sharks go after the ones leaving blood trails and that you're not one of them.

@Valay_17 (1/2) There's a video (I can't find it right now, so no link) demonstrating the need for child seats. In it, a bodybuilder tried to hold on to a child dummy in an actual 30 mph (50 km/h) car crash (obviously the guy was strapped in tightly, had a helmet, etc.). Even though he was curled up around the dummy as tight as possible, he wasn't able to hold on to it, and the dummy exited the car through the windshield. Now, if a bodybuilder can't even do that much, do you really think you're going to be able to hold on to anything at all in a 160 mph (250 km/h) crash?

@Valay_17 (2/2) Studies have shown that being in the back of the plane is, generally speaking, safer than being in the front. I think that's where you're getting the "run to the back" idea from. But, you see, those studies implicitly assume that everyone's safely strapped into their seats, not running around the cabin while failing to hold on to random crap. That is a much, much, much more dangerous place than seated with your seatbelt fastened, even if said seat is in the very front row.

@HiddenWindshield 1/2 - Yup got your point. 2/2 - Yes, that’s exactly where the “run to the back” idea comes from. I understand the concern of you guys explaining me that seatbelts are there for a reason and I do agree and I am not against them at all.

@Valay_17 In the miracle on the Hudson the back of the plane sank fairly quickly and the rear seats are farthest from the exit. Unless the plane noses down into the water I'd say the back is not the safest place to be. If the plane broke in half the rear would likely sink like a rock.

Downvoted because - in addition to all the good points made in comments - the question simply can't be answered unless you know the specific type of plane, and details of its construction.  E.g. I would imagine that something like a LongEZ, with its foam core construction, might float very well.


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What is the correct way for pilots to say the time?
While listening to Air Crash Investigations, the pilots used the 24-hour format in a different way than we do.
Example: 07:57 hrs would be said as oh-seven-five-seven hours (which makes it a little confusing as one might think of them as co-ordinates) instead of saying 7-hundred and 57 hours.
What is the correct way for pilots to say the time?

What are there hundreds of? Minutes?

@Harper-ReinstateMonica It is the hours as in 7 hours and 57 minutes

Even outside of aviation, most people would only say "oh-7-hundred" for 0700, where the "hundred" represents the double zero. I've never heard anyone add minutes after the "hundred".

@RobinBennett - exactly you just say hours at the end and minutes are never mentioned.

@Harper-ReinstateMonica - and Robin - surprising you haven't heard that one; it's pretty common, I think, to use the word "hundred" as the "hours marker", even if there is then a minutes (so in reality there's nothing that "looks like a one hundred").  I guess it's sort of vaguely "military style"?  If you think of it as a "mistake" it seems to be common

@Harper-ReinstateMonica No, it's hundreds of seconds, in base 60.

Like @RobinBennett says, I have literally never heard anyone say 24-hour time as "seven hundred and fifty seven hours". It would be "oh seven fifty seven" or "zero seven fifty seven", hundred is only ever used for the start of the hour as "oh seven hundred" or "zero seven hundred"

"Zero seven five seven" is the correct way to state the time, pronouncing each digit separately per the table below.

Aircraft call signs are sometimes grouped instead of annunciating each digit, for example United 6330 would be "sixty three thirty" instead of "six three three zero".
 Otherwise, headings, time, coordinates, and all other numbers used in aviation radio communication are generally spoken individually, digit by digit.
It is important to note that in aviation the need for clarity overrides common English usage.  For example, while it is common to pronounce the number zero as "oh" in the everyday world, aviators are taught that zero and O are distinctly different:  One is a number, and the other is a letter of the alphabet.  They should never be substituted for one another, or used in the wrong context.
Finally, the example you offered is incorrect because it isn't "seven hundred and fifty seven hours", past time zero, rather it is "seven hours and fifty seven minutes".  (Except it isn't said like that...)

"Aircraft call signs are sometimes grouped instead of annunciating each digit, for example United 6330 would be "sixty three thirty" instead of "six three three zero"." Maybe in FAA-land, but not in the rest of the world!

Shouldn't it be "fife"?

In practice, do pilots literally pronounce numbers as in that table above?  It's Tree like the plant and Fife like the instrument?  I've heard "niner", but haven't noticed "tree, fow-er, fife".

Actual pronunciation depends on radio reception. The short answer is no, for 95% of VHF radio transmissions pilots will say the numbers conversationally like everyone else. The exception is that nine is pretty much always "niner". (and zero is never "oh"...)  It is when transmission is weak or garbled that speaking needs to slow down, and each digit enunciated in a exaggerated manner per the table above to ensure effective communication. This can often be necessary on transoceanic flights, particularly when using older HF radios.

Nice. (i) Is there a reference for this? (ii) I can imagine pronouncing the non-word "FOW" in more than one way, like the word "FOE" for example. (iii) Is it really "NIN" "ER"? I've only ever heard "NINE" "ER". (iv) Is it really "TOU" "SAND" like the word "SAND", or is it more like "TOU" "ZUND"?

There may be more info in a radio telephony document, but I don't have a reference.  FYI, the table above is imperfect – and it doesn’t have all the Webster’s dictionary pronunciation symbology. However, I think you may be overanalyzing this. What is important is to understand the reasons behind these variations. Let’s look at a couple for example: Nine in German means “No” in English, so we add “er” to avoid confusion. (This one is used even when communication is clear) It is pronounced just like the number nine, plus “er”. (NINE-ER, not NIN-ER with a soft “i” as the table seems to show)

Regarding saying thousand, you would say the last half like you normally do. (It is isn’t important to emphasize “SAND” like sand on the beach, say it like you offered: “ZUND”.)  The important part is to emphasize the percussive consonant “T” over the soft sound of “TH”. Think about it, if you are trying to hear a weak broadcast on a static filled radio would you be able to discern a “TH” sound? No, because it sounds just like the static!

When transmitting time, only the minutes of the hour are normally required. However, the hour should be included if there is any possibility of confusion. Time checks shall be given to the nearest minute and preceded by the word ‘TIME’. Co- ordinated Universal Time (UTC) is to be used at all times, unless specified. 2400 hours designates midnight, the end of the day, and 0000 hours the beginning of the day.

(From CAP 413, but most regulations across the world are almost identical)
Examples:
08:23 = "Two three" or "Zero eight two three"
13:00 = "On the hour" or "One three zero zero"
07:57 = "Five seven" or "Zero seven five seven"

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

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Launch control equivalent while take-off
What is the proper term used when the plane is stopped and the engine is ramped up and after a certain thrust is reached, it suddenly starts off with a jerk.

The term I'm coming across the most is "static take-off" (even used by Airbus).
Its advantage is that it's using the least distance since you don't "waste" the distance it takes the engines to spool-up (in a jet anyway) to take-off power.
Not to be confused with "non-rolling take-off", which just means the aircraft was holding on the runway before the take-off run started. A non-rolling take-off is not a "static take-off" if the pilot didn't step on the brakes until the engines spooled up.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


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How to calculate position of a canard?
does anybody know how one can calculate the position, shape and size of an aircraft with a canard?
So the main goal of a canard configuration: increase lift + increase agility/maneuverability
So at least I can calculate the centre of pressure of the whole aircraft, which has to be behind the CoG to guarantee static stability in pitch.
But still, I have a range in x-direction , where I can place the canard. What do I have to define or know in order to place my canard?

Lift: I can design my main wing and its lift A_W_wing. In horizontal flight A=W=A_Wing +A_Canard ; So I could now design my A_Canard (shape, size).

But what about the maneuverability? What do I need to now?
So for my opinion: for a curve to fly with a specific rotating rate, my control surfaces (tail + canard) need to create a specific moment. Since Moment = Force * Arm  I could make calculate the needed Arms/ positon ?
Thank you for any tips on literature or calculations

This is really a quibble unrelated to your actual question, but not too sure about the agility/maneuverability part, in a configuration that is intended to never let the main lifting surface get close to stall.  I don't think the Rutan designs were more agile in the pitch axis than conventionally tailed airplanes of similar overall configuration, because of the limited pitch authority baked into the design.

Agility is possible with a canard (Wright Flyer), but practically requires computer control.  https://en.wikipedia.org/wiki/Canard_(aeronautics)#Control

The best canard is placed behind and not ahead of the wing. If you insist on putting it ahead of the wing, make sure its wake will flow above or below the wing in cruise.

@Helmut K. you may wish to work with roll and yaw rate as well to increase agility.  Possibilities include increasing control surface travel and or size.  These are your "destabilizers".  There may not be as great a need to destabilize the aircraft when they are not in use.

There are several factors to consider, and depending on your preferences there is not one ideal size and place for a canard. For a first rough concept look at similar designs and use the same canard volume. That is the product of canard area and the distance between its and the wing's neutral points, relative to the aircraft's reference parameters.
$$V_{Canard} = \frac{S_{Canard}}{S_{ref}}\cdot\frac{X_{NP, wing}-X_{NP, Canard}}{l_{ref}}$$
Collect as many canard volumes as you can from different but similar designs and use that for a first draft.
Next, decide on your priorities. Do you want good pitch damping for a pleasant ride even in rough air, use a smaller canard area on a longer lever arm. Damping grows with the square of the canard lever arm while its stabilising effect is only linear. Do the opposite if you prefer more agility.
Next, what center of gravity range do you desire? If, for example, you plan a tandem two seater where due to engine weight and pilot position the second seat is at some distance from the center of pressure, increase canard area accordingly.
So far we looked only at area and lengthwise position. But the vertical position must also be considered. Ideally, the wake of the canard will be above the wing for all practical angles of attack (from cruise to stall), which can be achieved by placing the canard wing high on the fuselage and the wing low. In that position, however, the canard wing might obstruct the pilot's view, so maybe you want to strike a compromise where the wake is below the wing in cruise but above near stall. This can only be achieved with a short lever arm, however.
Regarding the shape: A straight, high-aspect-ratio wing will be best. After all, the canard will fly at the highest lift coefficient of all surfaces so it should have the highest aspect ratio. Use a trapezoidal planform with a ratio of tip chord to root chord of 0.6 - 0.7, or, for simplicity, a rectangular planform which needs only one rig/mold for both sides.
But if what you really want is a good airplane and aren't determined on making it a canard: Place that smaller surface behind the main wing.

This diagram is from the book Design for Flying by David Thurston and might be of use.

Do you have a reference to this book (e.g. isbn)?

isbns here: https://www.amazon.com/Design-Flying-David-B-Thurston/dp/0070645574.  Be forewarned it's more of a "guide for laypeople" kind of book on light aircraft design and he only devotes a page and a half and that illustration to canards.


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Why does the moment coefficient for angle of attack = 0° not have to be negative?
I know for static stability reasons, the derivative of the moment coefficient of an aircraft has to be negative.
What I do not understand, why the moment of coefficient (cm) at all is allowed to be positive? Because when it is positive the aircraft pitches up.
Is it because when cm>0 I use my tails for trimming ? But is this "static stability" when I actively use my tail ?
I struggle with:
alpha = 0 ° --> e.g. cm=0.3    : the tail is used for trimming so that it counteracts the positive moment of the wing
alpha = 2° -- > e.g. cm=0.4 : now the pilot could also use the tail for trimming the higher moment
Thank you for your hints

An airplane achieves steady flight at a given airspeed only if the total pitching moment about its CG is zero. This is achieved with a particular fixed tail angle/elevator angle/canard angle. If the AOA corresponding to the total lift at this airspeed is 0 AOA, then so be it. It's just that for typical airplanes at operational speeds (away from $V_{FE}$ for example), the AOA is typically larger than zero; this means that, for a typical plane, the pitching moment at 0 AOA must be larger than zero at the tail angle, elevator angle, etc. corresponding to the trim speed for static stability.
Why larger than zero? Static stability means that when perturbed from the trim AOA (at which the pitching moment is zero), the resulting pitching moment will return the aircraft back to its original trim without pilot compensation. Take a look at the graph below; note that the moment is zero at AOA=1.5 (i.e. the trim condition).

If the pitching moment at zero AOA is negative (the orange line), then the airplane will diverge from its trim point when perturbed. That's the opposite of static stability.


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What is the sense of the Aerodynamic Centre for Design Process?
This topic has been discussed already several times, but I do not really get it. I think this one is a quite nice explenation: enter link description here
So I understand that the Aerodynamic Centre does not change with angle of attack. But what I do not get: how this helps in a pre-design of my aircraft?
Because: for stability calculation, I have to consider the forces acting on the whole aircraft and the corresponding lenghts (moments).
So regard a stationary horizontal flying aircraft: when I make the equilibrium of moments around the centre of the gravity, I have to regard the force F_cp,w at the centre of pressure of the wing (resulting lifting force) with distance l_cp (from centre of gravity) AND the force F_cp,t (resulting lifting force at the tail) with it's lever arm l_cp,t .
The balance becomes around the centre of gravitiy: F_cp,w * l_cp,w = F_cp,t * l_cp,t
When the angle of attacks change, F_cp,w*l_cp,w will change since the centre of pressure will move and the resulting lift will change.
So what I assume: there is a trick with which I can make the moment of equilibrium around the centre of gravity and where l_cp,w remains constant = Aerodynamic Centre
But I can not imagine how this is working with the balance of moments.
Can someone help? I am really confused.
Thank you very much

Maybe you want to surround your formulas with $ signs as in $F_{cp,w}$ etc. to make it appear as $F_{cp,w}$.

Aerodynamic Centre (AC) is particularly handy when we analyse the dynamics, that is, changes. For instance, changes due to disturbances: stability.
First thing, for analysis, we split all forces into forces acting where we want plus moments caused by the fact that these forces actually act somewhere else. And then when we write our balance equation, we need to balance both the forces and the moments.
We want forces to be at CG? Fine, let's declare that lift happens at CG. But then we acknowledge that apart from just lift, there will be an additional moment with respect to CG.
Now, due to linearity of aerodynamics at small angles of attack (AoA), this moment will change linearly with AoA. This effectively means that there will be a certain fixed point at which the moment will not change; or in other words, a point at which all the additional lift (due to pure AoA) can be thought to be happening. This point is the AC.
This is great, because:

We can express this extra AoA-related lift as $L_{w\alpha} \cdot \alpha$, where $\alpha$ is AoA and $L_{w\alpha}$ is the wing lift derivative by AoA, which is an intrinsic measurable property of a wing (or airfoil/aircraft).
The moment produced by this lift (with respect to CG) is that lift times its arm $x_{wac} - x_{cg}$ ($x$ is a relative distance (coordinate) of a point with respect to some datum, typically the front of the mean aerodynamic chord (MAC), which is another fixed property). So, our moment from lift will be $L_{w\alpha} \cdot \alpha \cdot (x_{wac} - x_{cg})$.

Remember though that for all but symmetric airfoils there will also be a static moment $M_0$ due to the camber-related lift also not acting at our CG.

In the above moment equation everything except for $\alpha$ is constant for a given wing/airplane. When we analyse longitudinal dynamics, which is all about AoA changes, we need the derivative by AoA. This leaves us with a constant, which unequivocally characterises our static stability.

Let's add the tail. It's lift will be trivial, $L_{t\alpha} \cdot \alpha_t$ plus some static lift at the chosen zero AoA. The arm of the AoA-related lift with respect to CG will also be constant: the distance between CG and AC of the tail. (Given that the tail airfoil is often symmetric, the whole lift arm is simply the distance to 1/4 tail MAC).
So, when we equate the moments for a straight-and-level flight (and ignore the downwash effect for a moment, which changes the tail AoA $\alpha_t$ from the base AoA), we get
$$(L_{w\alpha} (x_{wac} - x_{cg}) + L_{t\alpha} (x_{tac} - x_{cg})) \cdot \alpha = const \Leftrightarrow M_\alpha \cdot \alpha = const$$
The $const$ represents the static moments to be trimmed out; but what's important for the dynamics analysis is that we get a constant moment derivative $M_\alpha$: everything in the brackets is constant thanks to the 'constancy' of AC. We can break it up as $M_\alpha = L_\alpha \cdot (x_{ac} - x_{cg})$, which defines the overall AC of the wing + tail system, often called neutral point (NP).
Now if we want a statically stable aircraft, i.e. the one which negates the changes to AoA, we need, by definition, a negative $M_\alpha$ (and that's regardless of the trim $const$). From there, you can easily derive the requirements for the CG position with respect to AC. This very convenient, and this is why we need AC more than anything else. (In aircraft design, unlike pure airfoil/wing aerodynamics, CP practically never appears anywhere). Finding the AC is not always trivial though.
(Of course, in practice engineers use dimensionless coefficients rather than actual forces/moments, but I didn't want to introduce more entities than necessary here).

I will go through in detail your answer tomorrow Zeus - so far: thank you.What is more: when I understand it correctly: for longitudinal stability the CP of the WING is allowed to be in front of the CoG of the AIRCRAFT as long as my tail can balance the destabilizing moment of the wing, right?

Yes, this is a possible situation. For stability analysis, AC is everything, CP is nothing. CP in front of CG s not destabilising by itself; to see what is stabilising and what is not, you need to consider derivatives. A typical wing alone will be unstable, but a long tail will easily 'outmoment' it.


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Are there aircraft with multiple inlets for a turbojet engine?
Are there aerial vehicles which have multiple inlets with air ducts (shaped in S-forn) , which all lead the air to one single turbojet engine?
I have not enough space to put just one inlet area on my vehicle, so I thought about using inlets, located at the sides.
I think it is really difficult to combine the air of all inlets so that there is a uniform air flow in front of the turbojet -  because I don't have much space inside my vehicle.
Besides, I think is is really challenging when it comes to some maneuvers, where the flow sometimes will not flow through one of these inlets.
What do you think and do you know any vehicle from history?

Aren't almost all single-engine, multi-intake fighters fit this criteria? F-35, almost all Mirage's, Gripen, HAL Tejas, etc etc?

And the A-4, F-104, F-105, F-106, F-20,  U-2, ...

It’s common on military fighters. For instance most single engine fighters eg F-35, JAS-39, J-18, etc. will have a pair of inlets on either side of the fuselage feeding these single engine.  Some aircraft will have auxiliary air intakes eg B-2, F-117, F-22, MiG-29/35, SU-27/30/33, etc. to provide additional air on takeoff or other high power settings or to provide protection against FOD ingestion.

Quite a few planes fit your description of multiple (two) inlets for a (single) turbojet. The latest would be the F-35, other (as mentioned in the comments are numerous: JAS gripen, most Mirages, BAe Hawk, F-104 Starfighter etc.
I would say, without much thinking, that single intake jets with body mounted single engines are actually more rare. Even with a single intake, the air rarely has a straight path to the engine. Take F-16 for example, the air has to flow around front landing gear bay, and with MIG 21, the damn pilot sits in the way of a straight airflow.
The airflow to the compressor does not have to be absolutely uniform, the flow of air will quickly equalize across the compressor stages. The first stage might experience some extra stresses from uneven flowfield, but as the widespread use of multiple intakes shows, these stresses can be mitigated to more than acceptable levels.
As for maneuvers that would block airflow to the engine from one inlet: I don't think such maneuvers exist. A fighter jet can do a tail slide, and it's engine(s) keeps running. Even if the plane is going backwards.

Airflow is an issue mainly at supersonic speeds. It was a problem for the Sr-71. In too sharp turn or side-slip at cruise speed (M3.2) the shock wave from the shock cone (the sharp spike in the middle) would shift off the inlet lip, causing sudden drop in air flow and consequently engine thrust (called ‘inlet unstart’). But I don't think it affects any aircraft slower than M2.

Oh, true, but I think it's safe to neglect past mach speeds here

Darn pilots, always getting in the way! Also, "inlet unstart", what a lovely euphemism...

If you're designing a civilian aircraft, take a look at Diamond D-Jet.


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Why are some aircraft's life limited by both cycles and flight hours?
For instance the Pilatus PC-12 life is either 27,000 cycles or 30,000 flight hours, whichever comes first.
I understand that the limit life in cycles is mainly due to fatigue in terms of pressurization/depressurization, but what about the maximum flight hours?
Imagine a world where the aircraft would fly only one flight of 30,000 flight hours, then there would be only one pressurization cycle, and one takeoff, so what limits the aircraft to fly more than 30,000 FH?

Most of the components in an aircraft need maintenance based on the number of flights hours. I am not familiar with the Pilatus, but most probably a lot of components need to be replaced at the 30,000 flight hours mark making it economically speaking infeasible to restore it to flying conditions

I think some components of the airframe could experience fatigue with flight hours as well. Think wing flex with turbulence, for example.

Some fatigue related issues that are not directly tied to departures and arrivals, like wing bending from gust loads, but may be randomly continuous throughout the flight, need an hour limit.  Taking it to a comical extreme, if the airplane flew for 3 years straight without ever landing because it had a new pixie dust powered engine, it would have close to 26000+ hours of bending fatigue time on the wings and tail, engine mounts etc., but only one flight cycle.
Also, there will be an analysis that establishes the average amount of time per flight cycle, overall across the fleet, and this may be applied in some cases to give an hours life limit that is roughly equivalent to the target number of cycles, when applied in the aggregate.  That ratio of hours/cycles suggests that the average flight time per cycle for the PC-12 is somewhere around 1.1 hours.  In the Regional Jet world, the average flight cycle is about 1.3 hours.  The 30000 hour limit prevents a PC-12 operator who does a lot of long range flying with 2 hour+ trips from running his airframe up to 50000 hours before it gets to 27000 cycles.
30000 does't sound like a lot, but corporate aircraft generally live fairly pampered low-utilization lives and take quite a long time to get those hours, perhaps 30 to 40 years.  A regional jet gets to 30k in only 10 years.

You also get aircraft used in Skydiving whose cycle can be as quick as 15 minutes. The PC-6 used to be about 12 minutes to climb to 12000AGL and let the skydivers out; dive at the ground and level at 6000AGL; line up the runway and dive at the ground, level out and land. I once had to stay in the aircraft and boy, was that an exciting ride!


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When does a Maximum Taxi Weight Restriction happen on a 737-800?
This is the first time I experienced restriction in taxi weight on B737-800. Have you experienced that before? All I know that restrictions will be from MZFW/MTOW/MLW.
When does it happen?

MZFW = Maximum Zero Fuel Weight, MTOW = Maximum Take Off Weight, MLW = Maximum Landing Weight, MRW = Maximum Ramp Weight, MTW = Maximum Taxi Weight (MRW is synonymous with MTW) ... I think

@CGCampbell MTW and MRW are synonymous, or at least were up to my retirement in 1999. In the case of 747-100, 200, and 400 aircraft, the MRW was 3000lb greater than the highest MTOW. I say "greater than the highest MTOW" because the MTOW varied depending on the center of gravity.

The Maximum Taxi Weight (MTW) is one of the weight limitations in the FCOM:

(Boeing 737-800 FCOMv1 L.10.3 Limitations - Operating Limitations - Weight Limitations)
It can be the limiting factor when the actual takeoff weight including required fuel is very close to the MTOW and adding taxi fuel would get you above this limit. Note that the difference between MTW and MTOW is only 500 lbs (about 2 hours of APU use or 35 minutes of taxi). If the required taxi fuel is higher than that (which can happen at a busy airport), you are limited by MTW.


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Can airliners take off, cruise and land while unpressurized?
I'm wondering if the pressurization of the cabin has a part to play in the structural integrity of the fuselage, the same way an unopened soda bottle is stiffer compared to one that's open.

Airliners can still fly with holes in them, so even if pressurization plays a part in structural integrity, it's not a necessary part.

My hunch is that you are talking about jet airliners at 30,000 ft. But airliner is a broad term that can refer to a variety of aircraft, many of which normally fly much lower. In any case, takeoff and land are not issues at all - the pressurization of a typical jet is equivalent to several thousand feet above sea level - once you are below that physical height you would normally be at the "real" pressure. The real question is "cruise".

@GregHewgill https://en.wikipedia.org/wiki/Aloha_Airlines_Flight_243 is an even better example.

DC-3s were airliners, and didn't even have the capability for cabin pressure.

If leaving the cabin unpressurized is an issue, the airliner just sitting on the ground is a problem.

If an airliner experienced sudden depressurization at altitude, and then disintegrated during the emergency descent because of immediate structural failure, people would get upset.

Didn't MH370 flew quite a while after depressurization?

You are aware that near sea level, where the air plane experiences high loads during take-off and landing, there is no pressure difference to the surrounding air, right? That kindof answers your question, if you think about it. (The soda bottle has a much higher internal pressure than the surrounding air at all times.)

Why might they not, except if you were Asking about flying at (great) height?

There is no aircraft limitation on the 737 that requires pressurization to fly.  Realistically, even on O2 the whole time, going very high could be unpleasant.  I think the Air Force limit was FL 250 for routine unpressurized flight (recalling the T-37); when we depressurized the C-130 for high altitude air drops, we had some amount of pre-breathing that we did, and I never dropped anything near, at, or above FL 250.
But all of those are policy limitations, based on the human physiology involved, not the aircraft itself.
In airline life, I've done a maintenance ferry flight at 10,000' unpressurized, which is the highest we can have the cabin without needing to put on O2 masks.  There are probably ways to ferry higher than that if you really had to (crossing the Rockies, for example), but it wouldn't be preferable.  Again, though, that's looking at the human element rather than the jet.
I suspect that the reason that there aren't any limitations as postulated in the OP is that the aircraft itself is sufficiently strong that the extra rigidity of pressurization doesn't add anything necessary.  After all, the aircraft structure has to be able to withstand a sudden depressurization at cruise altitude, along with the ensuing emergency descent down to 14,000'.  When the structure is strong enough to handle all of that, I expect that what's gained by pressurization is just icing on the cake.

Additionally, aircraft take off and land unpressurized, so the fuselage needs to be strong enough for that anyway.

Actually, they takeoff and land very slightly pressurized, a few hundred feet worth, to allow the pressure to be tapered on and off gently so that passengers don't sense the "bump" in pressure.

And of course most people riding in a C-130 are at their peak - young, physically fit, with military training specifically for that sort of thing.  On a 737, you've potentially got people who are children, elderly, or with any number of physical ailments that would be a problem in an unpressurized flight.

With portable oxygen or flight deck crew only (flight deck oxygen has much more capacity than passenger systems) there's no reason a ferry flight with disabled cabin pressure couldn't go to FL 250.

As as aside, sometimes the cause of no pressurization can impose an altitude limitation, say, no pressurization due to two malfunctioning packs might cause an avionics cooling issue at too high an altitude, but that a secondary effect.

@nexus_2006  Good point - all packs inop may well ground the aircraft. But that's a separate (tho not unrelated) situation. The ferry flight I did was due to a suspect pressurization controller; we at least had good packs, thus adequate cooling.

Yes airliners can fly unpressurised, because pressurisation is not part of the critical load calculations.
The aeroplane fuselage is dimensioned using the critical G-loads from gusts. The largest gust impacts are at:

low altitude (higher air density, ground effect)
low airspeed (gust load is introduced more rapidly into the structure)

Both conditions are at take-off and landing, where the pressure differential is much lower than in cruise.

This actually did happen with a Boeing 737 on Helios Airways 522. Due to a bad switch setting and lack of communication between ground and flight crew, the pressurization was disabled. The plane sounded an alarm, but it used the same sound as an alarm that meant something completely different on the ground. The flight crew mistook that alarm for the one that couldn't happen in flight, and therefore ignored it. The plane never pressurized so the internal and external pressures were equal for the whole flight. The flight crew became incapacitated, the passenger oxygen masks dropped, and the plane flew its entire flight plan unpressurized on autopilot, up to entering a holding pattern over its destination. One interesting thing that did happen is that the alarm light for avionics cooling came on. The avionics are air-cooled, and without pressurization there might not be enough air to cool them. Even with the alarm, the avionics including autopilot continued to function.
Unfortunately the crew never woke up and never landed the plane. It eventually ran out of fuel and crashed, killing all aboard, but the plane itself functioned perfectly until it ran out of fuel.

Because of the cost of flying an empty plane, commercial flights are sometimes undertaken unpressurised, according to a British Airways maintenance engineer I knew (so it’s an unverified third-party claim).  They stay below 10,000ft - normal pressure altitude in an airliner is 8000ft based on my own measurements.


text:
What is the thrust to weight ratio of the A330-900neo?
Does anyone know what the thrust-to-weight ratio is for the Airbus A330-900neo?
I'm getting different numbers when I try to calculate this for myself.

Maybe tell us what numbers you have, where you found them and how you’re calculating it.

The thrust-to-weight ratio is not a single number because both thrust and weight vary throughout a flight:

The thrust-to-weight ratio varies continually during a flight. Thrust varies with throttle setting, airspeed, altitude and air temperature. Weight varies with fuel burn and payload changes. For aircraft, the quoted thrust-to-weight ratio is often the maximum static thrust at sea level divided by the maximum takeoff weight.

(Wikipedia: Thrust-to-weight ratio)
For standard sea level conditions with maximum takeoff thrust and at maximum takeoff weight, we would get the following numbers (source) for the A330-900:

Thrust / Engine
Total Thrust
Mass
Weight
Thrust-to-weight ratio

324 kN
648 kN
251 t
2462 kN
0.263


text:
Is it possible to power an ultralight airplane with a single-cylinder weed-whacker engine?
Plane:
Weighs 200 pounds.
16 FT wingspan,
36in. by 5 in. airfoil
Engine: 3,000 RPM, 2 HP
Propeller: 4ft with 4 blades and weighs 4 lbs.
Is there perhaps an equation that could determine this?

Related: https://aviation.stackexchange.com/questions/45160/can-an-ultralight-aircraft-fly-with-a-18hp-engine/58375#58375

Man, 2hp is really pushing it. Last time we checked, we kind of all agreed you need about 20hp to be safe.

Possible duplicate of Is it possible to fly using a small engine?

Don't worry about the weight of the propeller; it doesn't make any difference.

2hp will give you about 10 lbs of thrust, tops (it's generally around 4-5 lbs/hp).  If you have an ultralight with an all up weight of 400lbs, say, that maybe has a 10:1 L/D, you will need 40 lbs of thrust just for level flight at best L/D speed, requiring about 8-10 hp.  That's with no surplus power for, you know, taking off and climbing and such.
So a weed wacker won't get you off the ground in the first place in any kind of craft that can carry a human, unless of course the wings are like those on a human powered aircraft with a span that goes off into the next county.
And your ultralight has a small wing with only 48 sqf of area that will have a wing loading of 9ish psf and will need to go 50 mph just to fly, which it ain't gonna do on 10lbs of thrust.  Faggeddaboudit...
For an ultralight, you need wings big enough to get the wing loading down around 3lbs/sqf, to get a stall under 30 mph, and a thrust level equal to about 2 to 3 times the drag at max L/D.  If your ultralight has 40lbs of drag at max L/D, you will want to have at least 120lbs of thrust, or about 25hp minimum.  You can get by on less power if your ultralight is very clean, with long wings, like a Lazair or Mitchellwing, but not that much less.

There are methods of calculating the power required to fly a particular plane.  In your case, assuming that you'll also need to carry ~18 kg of fuel and a pilot (roundly 75 kg), the answer is NO.  A 2 hp (1.5 kw) power plant will be insufficient to give safe takeoff and climb performance -- with the possible exception of a high performance sailplane airframe (some self-launching sailplanes use ridiculously small engines).

My lightweight, high-performance, self-launching glider has a ridiculously small engine: 25hp as @JohnK predicts. No way it'll happen with 2hp! Just thinking about the definition of hp, we see that a 550lb frictionless, dragless airplane could still not climb faster than 2' per second, or 120fpm.

Here is how to determine the answer (which, as pointed out by others here, is no):
We assume you know the weight of your plane, its gliding airspeed, and the glide ratio. This means you know the sink rate in feet/second which multiplied by the weight of the plane, yields the power dissipation rate during a glide which can then be converted into horsepower.
To hold altitude constant then requires that the power application rate equal the dissipation rate. To climb requires that the power application rate exceed the dissipation rate by some convenient margin (your choice).

So power required = speed * weight / lift-to-drag-ratio (multiplied by some horrible constant for non-metric units, presumably)

Not sure. the answer is yes if the lift to drag ratio is exactly equal to the ratio of the forward speed to the vertical speed but that is an assumption I do not feel qualified to justify. Peter Kaempf on the aviation stack exchange would certainly know the answer.

@RobinBennett   Yes!

"Is it possible to power an ultralight airplane with a single-cylinder weed-whacker engine?" -- I've seen someone actually try it. The engine was still attached to the weed-whacker. The string was replaced by a propeller. The unit was strapped to the pilot's waist, projecting backwards. The aircraft was a paraglider. The pilot foot-launched off a mountain and then started the contraption in the air. Unfortunately I never heard what sort of climb rate he was able to achieve once he flew away from the ridge lift.

It would give him a flatter glide.  A bit flatter.  Paramotors get by on a rock bottom minimum of about 100 lbs of thrust and most are 140-160.  2hp would give you about 10lbs tops and would require a reduction drive.

For comparison, the Wright flyer had a 12 horsepower engine and the minimum requirement by the Wrights was 8 horsepower. That's not to say there can't be a smaller/lighter plane, but they definitely started small and slow, by today's standards, and the resulting numbers are very similar to the other answers of 8+ horsepower.

In theory, yes. Non powered gliders can fly with no HP just off of thermals, so a 2HP engine would be a reasonable boost for a plane like that. But to take off, that likely wouldn’t cut it.


text:
What's the maximum range an A350-900 can reach at TOW of 250,606 kg and full payload?

When the TOW (takeoff weight) of an A350-900 is 250,606 kg (because of runway length limitation), with maximum payload, what is the maximum range it can reach using the chart above?
It is easy to figure out for a Boeing 787 since Boeing provides different brake release slopes, as shown below, unlike the A350 chart.

Sorry but whats not clear in the chart?

Since Airbus does not publish this in the charts, we have to calculate it ourselves. You are asking about an Airbus A350-900 with a TOW (TakeOff Weight) of 250,606 kg. This weight would then consist of:
$$ \text{TOW} = \text{OEW} + \text{Payload} + \text{Fuel} = 142 \, 400 \, \text{kg} + 53 \, 300 \, \text{kg} + 54 \, 906 \, \text{kg} $$
based on typical OEW (Operating Empty Weight) and max. payload from Wikipedia. This leaves us with about 54.9 tons of total fuel. Not all of this fuel will be usable however, since the aircraft still needs some fuel reserves: $ \text{Fuel} = \text{Trip Fuel} + \text{Reserves} $.

To calculate the range of the aircraft, we can use the Breguet range equation (see e.g. this answer):
$$ \frac{\text{TOW}}{\text{LW}} = \exp \left( \frac{R \cdot g \cdot b_f}{v \cdot L/D} \right) = \exp(f \cdot R) $$
We are only interested in the TOW, LW (Landing Weight) and the range $R$, so I combined all the other aircraft specific factors into one factor $f$. The landing weight can be calculated via the fuel use:
$$ \text{LW} = \text{TOW} - \text{Trip Fuel} = \text{TOW} - (\text{Fuel} - \text{Reserves}) $$
We now have to solve for the two unknowns $f$ and the reserve fuel. For this, I read off two values from the plot you provided:

We now have two given ranges for known masses:

At MTOW and max. payload (resulting in 84.3t of total fuel) the range is about 5800NM.
At MTOW and max. fuel (about 112.6t based on about 25t of payload) the range is 8600NM.

By plugging these values into the Breguet equation, one can numerically solve for the unknowns:
$$ f \approx 5.2711 \times 10^{-5} \; \text{NM}^{-1} \;\;\; \text{and} \;\;\; \text{Reserves} \approx 10\,545 \, \text{kg} $$
We can test, if these parameters make sense, by plugging in no payload at maximum fuel:
$$ R = \ln \left( \frac{\text{TOW}}{\text{LW}} \right) / f \approx 9698 \, \text{NM} $$
This range is consistent with the value where the magenta line crosses the range axis.

Now moving back to your numbers. We can now plug in a TOW of 250,606 kg at maximum payload. The trip fuel will then be:
$$ \text{Trip Fuel} = 54 \, 906 \, \text{kg} - 10 \, 545 \, \text{kg} = 44 \, 361 \, \text{kg} $$
This gives the final answer:
$$ R = \ln \left( \frac{\text{TOW}}{\text{TOW} - \text{Trip Fuel}} \right) / f \approx 3696 \, \text{NM} $$


text:
How do pilots avoid thunderstorms at night over the pacific en route to Australia where there is no radar? And what if a large front develops?
While there is some limited radar on board for pilots, we hear that there is no radar over the ocean which is why flight locations are estimated. As the distance between the United States and Australia is lengthy, how do pilots avoid thunderstorms over the Pacific en route to Australia at night, and what did they do about them If a large front develops in their way?

Look out the window and see if you see lightning?  Even in a small plane, it's possible to see thunderstorms ~100 miles away, say storms in west central Nevada from over the Sacramento area.

Keep in mind that route is dotted with islands, and many have runways and weather stations.

The on-board radar is not some limited backup tool, it's the primary tool whenever you have it, even if there is coverage from ground radars.

You get by on your airborne weather radar which is good out to about 80 miles.  You may also get help from other pilots ahead of you on your route providing PIREPS to ATC on HF, or talking to you directly if you are close enough to use VHF between each other.  But as far as picking your way around cells, besides being able to see them if you are in the clear, your own weather radar is all there is.
The main limitation of airborne WR is shadowing due to signal attenuation in rain, where the really nasty stuff is hidden (looks black as if the sky was empty) behind less nasty stuff.  Even with the modern radar there is a bit of skill and experience required to interpret the returns and not get fooled by shadows.

@ymb1: And why couldn't commercial planes use satellite internet to get info from weather satellites while en route?

Just to emphasise, the point is that the plane has its own weather radar set. It's not reliant on somebody else's radar to know where the rain is.

Internet based weather radar is not real time and there is enough of a delay in presentation to be a problem.  Pilots flying IFR in the US using an internet based weather overlay on their flat panel displays as if it was airborne weather radar have gotten into serious trouble because cells had moved significantly from where it was depicted.

@jamesqf satellite weather is only good for getting an overall picture of a system.  As far as non-airborne sources go, you need ground based doppler radar to see exactly where cells are within a weather system, which is what you are really concerned with.

@John K: It's quite possible to detect lightning from orbit, even geostationary orbit, in near real-time: https://www.nesdis.noaa.gov/content/tracking-lightning-space-how-satellites-keep-you-safe-during-thunderstorms  Of course this doesn't detect all weather, but at ~30K ft I think you're above just about everything but thunderstorms.

@jamesqf but you need to be detecting water not so much lightening, and you need to do it with precision, within a mile or two.  The only other weather issue that gets you into trouble, except when close the ground, is icing and WxR doesn't help you there in the first place.  IFR flying is a risky business when conducted w/o WxR if there is any convective activity on a flight.

@John K: I don't follow your reasoning here.  Thunderstorms have lightning, by definition.  (I'm not saying that there aren't other kinds of weather you might want to avoid, but that's not what the question is asking.)  Nor do I see why you need to locate T-storms within a mile or two.  Certainly I'd be much happier giving them a wide berth, since (at least at the SEL level) you can get turbulence &c many miles from the actual storm.

The protocol in the airline business is to keep about 20 miles away from the edge of a cell, preferably upwind because hail tends to be toward the downwind side.  To know that you want at least want to be sure of being that distance +/- a couple miles.  The other thing is the worst part of the storm, to avoid at all costs, is the rain shaft, the core in other words. Lightening can be happening quite a distance from that, so knowing where lightening is without knowing where the water is is a problem. Lightening strikes are far less of a hazard than flying into the rain shaft.


text:
What parameters to focus on when constructing a wing for turning?
I want to construct a flying wing (glider) with the sole purpose of performing a helical turn (or a circular turn with no propulsion attached) with somewhat of a fixed radius. What should be the main parameters of the flying wing that I should focus on?
Thanks

Is it okay if your aircraft is only good at turning one particular direction?

Strength, ability to withstand the G forces of turning.  Secondly, having a stall Angle of Attack as high as possible.
A low aspect ratio wing with a very thick airfoil is a good candidate, such as the ones designed by Hugo Junkers and seen on Fokker planes of 100 years ago.  These work very well for slow, turning flight.
Another candidate would be the delta.  Strong and easy to construct, deltas offer a very high stalling AOA, making them very good at turning tightly.
Deltas have more drag for a given amount lift than a straight wing, so the first candidate will stay in the air for a greater amount of time, the second may give a thinner helix.
Good reading articles can be found in Air and Space magazine:  "What the Red Baron never knew" and "The Perfect Airplane Wing".


text:
is it necessary for the velocity distribution profile over a flat plate to have that kind of parabolic shape?
A wind tunnel boundary layer experiment was done for a flat plate with 5 stagnation probes distributed horizontally (each probe has 5 vertical sensors) and a variable flap, for 2 different free stream velocities "V=20 m/s" & "V=25 m/s" and 2 different cases:

without the variable flap open;
with the variable flap (the angle is approximately 25°).

A picture of the test setup is shown below:

A schematic of the setup:

The pressure values over the plate were collected from the sensors, then the local velocity was calculated using $$\Delta P = \frac{1}{2} \rho v^2$$ and the 2 graphs below were plotted for each case:

However, as you can see in the case of a flat plate with a variable flat, I got a negative pressure value at sensor 5, which in return gave an undefined velocity value (you could see the sudden drop).
My questions are:

What could be the reason for that negative value?
Is my velocity profile okay? Or is it necessary to look like we all used to see (approximately a parabolic shape) because the one we always see is a curved line towards the top when mine is curved towards the bottom?
How can I tell whether I am on the laminar boundary layer or the turbulent boundary layer based on the graphs? (Or must the Reynolds number be calculated in order to find out?)

Do you have a picture or a schematic of the assembly?

@sophit for the moment I don't, but I can provide it, if that helps, the experiment was done in the TecQuipment AF1300 subsonic wind tunnel.

A simple schematic is enough, just to understand the setup of the experiment

I couldn't really provide what you asked for since doing a schematic isn't my strong suit, so I hope this helps so you could help me, here're some pictures for the setup: https://ibb.co/y5H45R2  & https://ibb.co/r669FbF

I would need a diagram of this experiment apparatus to see what’s going on.

@I.ham I added the picture you provided in the question, am I correct to assume that the pressure sensors are number in increasing order, from front to back?
So sensor 1, lowest 1, then 2,3,4,5 top of location one, and then the sensor downstream with lowest 6, 7,8,9,10 top of location 2?
Also, could you provide the raw measured data?
With some more information I can make a quick sketch so I can confirm we have the
correct understanding of the data and the setup

@ROImaisonThank you a lot you did great, here are some other pictures with an explanation, I hope it helps more, this is a schematic of the probes https://ibb.co/hfTbkqs, and this is how the probes are distributed https://ibb.co/7bjmvxd, this is the data for the first case at 20m/s https://ibb.co/3CV3cSH, and this is the data for all the cases (respectively open flap at 20m/s... at 25m/s then the last two with the flap at a certain angle at 20 m/s ... at 25m/s) https://ibb.co/DLQKD8X, note that Pt is the total pressure and P0 is the reference pressure at a certain point in a fluid flow.

1- what could be the reason for that negative value?

Probably a bad sensor. If it were me, I would re-run the experiment but swap two of the probes. If the negative value follows the probe, then it's definitely a bad probe.  If it does not follow the probe, then maybe something else more interesting.

This is lab work in college, so I can do it just once, but I still could check with the others; if you see both graphs, the one without a flap didn't have such a problem,  so I think if the sensor was bad it would show in both graphs, correct? (because it was the same plate only a flap was added).

It seems we need to go back to $\Delta$P.

is this the difference in ambient P and lower pressure created by the freestream?

This is what would be expected from the "aspiration" effect of a moving fluid on a pressure sensor.
Your first plot doesn't look like the others.  After checking the apparatus to make sure nothing is wrong with the sensors, you may have found ... an area of stagnation, made stronger by the flap.
Your $\Delta$P is based on the effects of freestream velocity.  A "negative velocity" calculation is actually ... an area of compression, also known as a high pressure stagnation zone.
Is this not what you are looking for?
Data from the "no flap" run seems to show a reading of lower velocity at both 20 and 25 m/s.  Make sure, in your calculations and presentation, that your "$\Delta$P" (and $\Delta$V) is based on either ambient pressure, or freestream pressure.

Looking at the setup, my guess is that the probes are mounted on the protruding elements and are perpendicular to the flow, and are all measuring stagnation. The difference between total pressure and ambient would then be used to calculate the velocity

It just seemed illogic for a stagnation zone to appear so soon, nearly at the leading edge, this is a sample of the data with no flap at 20m/s https://ibb.co/3CV3cSH the data was collected directly from the computer that's why I had no idea what was going on exactly and what went wrong? so I am just trying to have an explanation, and since I don't know how a correct experiment would go I couldn't make any comparison.

@I.ham. so the "odd" data was from the front of the plate, 5 being closest to it.  Angle of Attack (of the plate) will also have a lot to do with your results.

@I.ham, as the probes are protruding out of the plate, the flow will collide with them and stagnate.


text:
Is it possible to increase available take-off distance for a seaplane by turning?
In the video below an A5 had an accident after takeoff. The final cause of the accident seems to be a sharp increase of AOA above stall.

Do sea planes have an alternate technique for takeoff from tight spaces. Let me draw what I mean:

So in case 1 take distance available for T/O is not enough. Is there a procedure such as in case 2, where you use one side of the lake to accelerate to some speed and then turn to main leg to complete the takeoff?

That video is an example of poor airmanship.

Yes when I was bush flying in Northern Ontario (C-185 on straight floats) I used turning takeoff runs from time to time when loaded and with limited space.  You start the turn once on the step.  Step turns are a normal skill on floats although it's not a good idea to do it on choppy water in high winds.
The guy in the video had insufficient space to take off even with the turning takeoff run and it looks like he couldn't be bother to head farther down the lake to get a better into wind run.  I wouldn't have done a turning takeoff like that in that sort of windy chop.  I would've started right at the far shoreline as far as I could go, and determined a go-no-go point using a landmark that was about 2/3rds of the total run.  If you're not airborne fairly quickly (by water takeoff standards) in that kind of wind and waves, something is really wrong or your airplane is grossly underpowered.  I suspect that guy was overweight.

The mushing descent near the end made me wince

One prob is you have subsiding air flowing onto the water from the shoreline because of a bit of downhill slope effect which can take a couple hundred fpm off your rate of climb below 300 ft from the surface. I would get airborne and stay right on the surface to gain extra energy and speed margin and zoom climb through the subsidence. One of the many hazards most land pilots never have to deal with, beyond boaters etc.I used to do float ratings and always started the first briefing with "Floatplanes are easy to to learn. It's a big stable catamaran. The HARD part is staying out of trouble."

I can't see how the pilot thought that was going to work. They could have gone to the far end of the lake and started the run, doing a much more gentle turn to gain a bit more of a run. Instead they put themself into a situation where they had to do a tight turn immediately to avoid trees. Sad.

One of the biggest challenges (and satisfactions if you thrive on living-by-your-wits ambiguity) is dealing with the "shades of grey" of bush flying as opposed to go-no-go numbers you can rely on in normal flying.  All up weight when loading in a remote place?  Educated guess.  Takeoff distance available?  Looks long enough; try it, but be ready to bail.  Shoals and deadheads?  Keep a look out (I ran aground on a shoal once while taxiing).  Boaters wanting to race you?  As long as they stay far enough away but keep an eye.  Weather?  A mile and 500 ft ceiling can be sufficient for some trips.


text:
Why does the left-backswept wing have more leading edge suction on an oblique wing?
In this paper, it states:

The leading edge suction of the left backswept wing is larger than that of the right forward-swept wing at subsonic speed, and the lift vector of the left wing inclines forward (see Fig. 2(b)), which results in smaller drag of the left wing.

What causes the left back swept wing to have more leading edge suction? For clarity, this is an obliquely swept wing:

What is different about the right forward swept wing compared to the left back swept wing? Why does the left side have a higher amount of leading edge suction?
It also states that at subsonic speed this is true, which to me, hints that this is false at supersonic speed.

That's a very good question. If the leading edge suction is related to lift then, given the spanwise flow doesn't participate to lift, the RH wing should generate more lift (the leading edge is "more perpendicular" due to the sweep angle, so less spanwise flow). This is obviously wrong, I'm curious for the explanation.

The paper shows an image for the chord-wise pressure distributions for an oblique wing (Fig 2a), which indicates how the pressure distribution is modified due to forward ('right wing') and rearward ('left wing') sweep:

In case of the forward swept wing there is a reduction in suction for the front section, and an increase in pressure for the rear section.
This effect is caused by local boundary layer thickness modification due to the spanwise flow.
Important in this is to note that the flows are in opposite directions. In the case of rearward-sweep, the spanwise flow is from root to tip, whereas in forward-sweep the spanwise flow is from tip to root.
This answer shows the effect of rear-ward sweep on the boundary layer. It shows that the crossflow helps to 'thin' the boundary layer.
As Vos and Faroukhi - Introduction to Transonic Aerodynamics explain on p.470:

We should recall that the boundary layer thickness at any point on the surface is a function of the distance that the flow inside the boundary layer has traveled from the leading edge (see the discussion on p. 305). With that notion, we can easily deduce that the boundary layer over the outboard wing is thicker than what would be expected based on two-dimensional boundary-layer properties. Conversely, the crossflow in the boundary layer causes the thickness of the boundary layer at the root to decrease compared to its two-dimensional counterpart.

In the case of the rearward-swept wing, we have spanwise flow from root to tip, this reduces the boundary layer thickness which accentuates curvatures thus leading to a higher peak.
In the case of the forward-swept wing, we have spanwise flow from the tip to the root, this increases the boundary layer which reduces the curvature thereby spreading out the lift.

Sorry, forgot to paste the part of the paper that relates lift to leading edge suction

@mins I realized I misunderstood some text in the original paper. I rewrote the answer to focus on the chordwise distribution of the effects.

I’m sure I’m misinterpreting something, but why does the cross flow thin the boundary layer? I thought any flow that wasn’t exactly perpendicular to the chord would thicken the boundary layer? (I did read the linked answer) - Edit: I just re-read the quote in your answer, and that made sense, but why does the cross flow at the root thin the boundary layer?

Thickening/thinning depends on the direction. In general, there is an obstruction in the center that prevents crossing from the other side (either a mirror plane or the fuselage). If inboard flow drives particles to the centerline, these particles cannot go anywhere, resulting in an accumulation of air particles; a thicker boundary layer.
Conversely, in the case of outboard flow, the air particles are driven outboard. They cannot be easily replaced and the boundary layer becomes thinner.

Thanks for your answer. I've some difficulty to understand the paper: "The aerodynamic load of the right forward-swept wing is concentrated on the wing root; thus, the leading-edge suction of the right wing tip and the lift coefficient decrease". Let's assume the load is redistributed to the root, still the "thus" in the citation is not obvious. I can understand this changes the wing moments, but that doesn't explain why the lift coefficient and/or the LES coefficient decreases overall. Why would the lift at the tip "count" more than the lift at the root?

@mins I had a similar issue when I read the quote. I think they mean the trailing edge when they say "root". Then it would make more sense. In my opinion, one does not cause the other, it's all happening due to the crossflow.
Like I said in the answer, the boundary layer thickens, so the curvature of the airfoil decreases, leading to a reduced lift & reduced peak. Presumably, the effect is less present at the rear, as the curvature is already lower, leading to a relative increase.


text:
Why does changing the Re number increase the critical angle of attack of a wing?
In this answer, it says:

Normally, increasing speed reduces the impact of viscosity (expressed as the Reynolds number). Viscosity is the reason why airflow detaches, and less of it means that flow will stay attached to slightly higher angles of attack.

I'm a bit confused as to why the Reynolds number changing (increasing, I think) would increase the critical angle of attack (cAoA) of the wing. From what I've heard, the Reynolds number just represents the ratio of inertial to viscous forces, determining whether the flow will be turbulent or not.
Why would simply changing the Reynolds number affect the stall AoA?
One thing I just realized while writing this is that it could be due to the turbulence from the higher Reynolds number helping the flow stay attached (which would make sense to me). However, the way the quoted text above is written makes me think it's due to something else.

One of the links in that answer also led to this answer, in which it states:

The lift curve slope increases a bit with Reynolds number

This is also puzzling to me, but it might require a separate question.

Increasing Re (e.g. by increasing velocity) moves the transition region aft, therefore an increased AoA is needed to move it forwards (the critical AoA is when the separation is too much forward). Basically this is this question: What is the effect of airflow speed on separation? (again).

@mins Thanks for your comment. So basically the increased speed will help accelerate the air that has been decelerated over the wing? (At higher AoA the air will decelerate more over the wing than at shallower AoA) Also, this now faster moving air will resist separation more, allowing higher AoA

There are multiple effects at work. A higher velocity re-energizes the boundary layer by mixing slow and fast air in eddies, preventing low (or inverted) air velocity and separation.  Viscosity is also a cause of air velocity reduction. A larger Re means less viscosity efficiency (relatively to inertial effect). Both effects make it difficult to trigger separation.

@mins I see. So the higher Re will prevent the slower moving boundary layer air from separating by 'evening out' the velocity profile along the boundary layer? (higher Re causes more mixing of flow, evening out the velocity profile)

Correct (to prevent any ambiguity: "Re will prevent the slower moving boundary  [...]" means "Re will prevent the now faster moving boundary [...]")

@mins Oh okay, that makes sense, thanks. One last question: In the last part of my question it asks why the lift curve slope increases with Re number. Is it due to the same effects you just mentioned?

I cannot answer, perhaps this is similar to extending a flap, the downwash may be increased.

Bottom line upfront is higher Reynolds number will result in a thinner boundary layer with a fuller velocity profile. With the no-slip condition for the fluid over a solid surface (airfoil), the velocity will be zero right on the surface, then gradually increasing with distance from the wall in the boundary layer. Separation is defined as when, at any point in the boundary layer, the local velocity is zero. Then, in a more separated condition, the velocity reverses in direction.  Point being, all else the same (in a subsonic, 2D flow over an airfoil) at higher Re, the velocity profile is further away from these conditions to begin with than at lower Re.
What typically causes separation in this problem domain is an adverse pressure gradient and it’s important to note the circularity involved. That is, pressure gradient effects the boundary layer and details of the boundary layer effect the pressure gradient.
Some helpful references include:

Fundamental Mechanics of Fluids by I.G. Currie, 1974, section 9.11
Boundary Layer Theory, 7th Edition, by H. Schlichting, 1979, section 22.c


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Why is a conductive nose a desirable feature for stealth aircraft?
Recently I saw this illustration that lists the desirable characteristics of a stealthy plane.
Most of them make sense but I did not understand how a conductive nose could help the plane be more stealthy.
The conductive canopy part I searched and I got some conflicting information but overall that seems to be simple.
What's this conductive nose for? Wouldn't that make it more obvious?

Source

Not sure why this was down voted, seems like a reasonable on-topic question.

Do you have the source of the picture?

It's a prototype diagram from here https://www.secretprojects.co.uk/threads/blitzfighter-bushwacker-atlas-lightweight-attack-studies.9179/

Atom ionization occurs naturally at the nose due to air friction. It also occurs at any location directly facing the airflow, like the leading edges. A ionized material behaves like a shield for electromagnetic waves:

Radar waves from the outside are reflected to the outside. This makes the tracking by hostile radars easier by reinforcing their echos.

Radar waves from the inside are reflected inside. This prevents the aircraft from hosting its own radar antenna in the nose.

To hide from external radars and/or to use the nose as a radome, ionization has to be eliminated. In both cases this is done using a conductive material.
Details follow.

Radar is based on wave bouncing
Radar is based on wave reflection/refraction, detected as an echo. Materials with better reflective/refractive properties, like ionized ones, create stronger echos. Hiding from radars means not using such materials.
Ionization by friction, de-ionization by current
When the aircraft moves through air, electrons are ripped off atoms by friction (material and air atoms), leaving ionized atoms at the material surface and in air around and free electrons. De-ionizing atoms consists in making atoms neutral again. It's done by capturing free electrons circulating in the materials.
But free electrons, which is nothing else than what we call "current", move only in conductive materials, hence the material subject to ionization must be conductive to allow the circulation of free electrons and their recapture by ionized atoms.
Aircraft areas more subject to ionization are the ones directly facing the airflow, leading edges and nose.
Conductive radome material
Nose is often used for the radar antenna. The material here must allow radar waves to be radiated and received, preventing pure metal to be used, but at the same time must allow electrons to circulate, i.e. it must be conductive. The material used for this part, the radar dome, or radome, is actually high technology, there are many electromagnetic constraints to be taken into account. For more, see Why do Russian Su-30SM fighters have white nose radomes?


text:
Question about airline routing with oceanic flightplans
so I looked up my recently flight from Incheon, South Korea to JFK on Flightaware (Korean Air 85 on Wednesday 8/23):

Image source
The "route" under "Flight Data" is just 4348N/08224W IGN IGN1; however, my flight seems to take on a much more intricate route; on Skyvector, it looks like we went on airway Y697-G597 and Y51 between Korea and Japan, and then onto an oceanic routing that didn't really resemble the Pacific tracks, which usually are a lot more south than the route we took. Our route lead straight into northern Alaska. In contrast, this flight, also from Incheon to JFK, seems to include all of its waypoints and airways on Flightaware:

Image source
This BA flight from Heathrow to JFK also seems to lack detail in its flightplan. The first waypoint listed is just offshore Canada; there are no waypoints in the UK that guide it toward the North Atlantic Track system:

Image source
On US domestic flights, the flightplans on Flightaware seem sufficiently detailed, and the planes generally follow those routing. So my question is: on flights over the ocean, is it common for flight plans to include rather vague information on routing when filed initially (or could this just be an issue with Flightaware)? Would it then be up to ATC to issue an initial routing when the pilots pick up their IFR clearance? If so, I wonder why this would be the case, instead of the airlines' dispatchers just supplying their own routings.

FlightAware pulls much of its information, including routing information, directly from the FAA—and possibly from other air navigation providers in other countries. The routing that is stored in the FAA's computer will be what is shown in FlightAware.
When the route does not have to be updated, the original route as filed/processed will be retained in the computer. Your KAL81 example shows this.
However, when a controller updates the route of an airborne aircraft, the route information prior to the update point is no longer necessary; the computer throws it away, retaining only the update point and the new downstream routing. This is what happened to the KAL85 flight that you were on.
Note that the update point for KAL85 is 43º48'N, 082º24'W. Checking SkyVector we see that this point is just within Cleveland Center airspace. But compare to the FlightAware track: Your flight did not pass so far west and south, just off the eastern edge of Michigan. Instead it entered the Lower 48 in upstate New York, well within Boston Center airspace. So I'm not sure exactly why a controller updated the route using that coordinate point... but they did, and that is why FlightAware doesn't show the entire route.

Nice answer! Is the FAA source of that routing info available to the public, or only to providers like FlightAware & such?

Aha, thank you so much! Yeah, I remember shortly before boarding, the first waypoint was POWAL, which is far into the ocean--so I guess only the routing at and after POWAL was updated before I looked up my flight on the website. And then during my flight, it was updated once more

@RalphJ You can look through the SWIM registration process and see what the application form looks like. I imagine that it isn't available to Joe Q. Public, or at least the information would be heavily filtered.

@randomhead I think I have a new project to learn about... thank you!!!


text:
Do El Al Israel aircraft have Chaff/Flare dispensers and Radar Warning Receivers?
I have heard that El Al Israel aircraft have missile warning systems and lasers to stop anti-aircraft weapons. Do El Al Israel aircraft have chaff/flare dispensers and Radar Warning Receiver (RWR) too?

Related: Why don't El Al 747's stay at the gates at JFK? and How common are missile defense systems on civil airplanes?

From what I could find, they used to use a Doppler radar to detect a missile and launch flares to counter it in a system called Flight Guard but it was deemed too dangerous and a fire hazard so they replaced it with a new system. What they have instead now is a MAWS (Missile Approach Warning System) that is designed to detect missile launches very early on with methods such as an IR camera and it seems to use some kind of a laser to counter the missile by messing with its sensor and making it go off course. This system is called C-MUSIC
Here is a video with some... crusty CGI that gives a basic demonstration of how it works

"Climb flight level 3000" - wow, thats some altitude! You would think that the script of even a faked aviation video would have some clue about aviation terminology :D

I mean climbing to flight level 3000 would be a pretty effective way to avoid SAMs!

If you want to know if a specific plane has missile countermeasures, you could take a look at the rear of the plane. If there is an active countermeasure installed, you should be able to see a "box" mounted under the plane (pic attached).
FedEx tried to propose for the installation of countermeasures but the FAA halted the certification process in Jan 22 [Reuters]

EL AL are reported the C-Music IR counter measures system. It is not flares, but instead an IR strobe to counter the simple IR seeker heads in man portable IR missles.

Is there a link available to confirm what you posted? I agree that a jammer or DIRCM would seem to make better sense than flares, given the available technology.


text:
Why do Belugas keep executing a go-around?
I live under the approach path for the Airbus facility in Hamburg and very often a Beluga executes a go-around. It is quite loud since when it pitches up the engines are pointing at my house. I wonder why this happens so often, even when it is not windy outside. What factors cause such frequent go-arounds?
I have an image here showing two failed attempts at landing in a row from a Beluga, which I find quite astounding!
Edit: Link to complete flight data of presumably the same flight on FlightAware

Comments have been moved to chat; please do not continue the discussion here. Before posting a comment below this one, please review the purposes of comments. Comments that do not request clarification or suggest improvements usually belong as an answer, on [meta], or in [chat]. Comments continuing discussion may be removed.

I wasn't on the flight deck, nor do I have inside information, so this answer is an educated guess.
I assume this is flight-training for new Beluga crews and/or license re-validation which is included in normal line operations for the sake of efficiency.
There are no spare Belugas, so if new pilots are trained or some kind of license re-validation is needed (e.g. recurrent training or training after prolonged leave due
to illness / maternity/paternity-leave), training needs to happen on those aircraft that are used for day-to-day transport operations.
To minimise the impact on the transport schedule, it makes sense include the training in scheduled transport operations, for example by including a few go-arounds at the destination airport. This way, training has the least impact on schedule and it keeps the cost of training low.

The craft passes extremely low over us, only a couple hundred metres, and it's a very busy area of the city that is being disrupted. Why would it do that if it is just training?

@JoeBentley unless your city has noise ordinances that will fine them for unnecessary go-arounds, they aren't going to care about disrupting you. Or at least, not care enough to cost the company money by doing it somewhere else.

@JoeBentley because "just training" is a very important part of safely operating an aircraft. It needs to be done at an airport, and Airbus happens to own the airport and is virtually the sole user of it. So without causing extra fuel costs for ferrying the aircraft to a training airport, and not interfering with other (airline) operations at the airport, it seems a good choice.

@Joe Bentley - I picked one Beluga at random F-GXLI and looked at Flightradar24 data for it. Between July 20th and October 2nd it landed at Hamburg 21 times. Two of those landings included go-arounds, a flight from Madrid on September 21st and a flight from Saint Nazaire in France on October 2nd which landed at 2:14 pm CEST, right around the time that you posted your question, so I'm guessing that particular go-around prompted your post?  I think they fly four or five of them, so based on just this one dataset I estimate this happens about once a week, does that sound about right?

@Joe Bentley - I missed your "once or twice a fortnight" comment that you posted while I was searching on Flightradar24.  So it sounds like my once a week estimate was pretty close.

@JoeBentley Noise pollution isn't penalised, so companies won't care about the adverse effects on the very busy area of the city.

Even if noise pollution were penalised, it would still be a tradeoff: does the fine cost less than going somewhere else?

As a fellow Hamburg resident, they do come in pretty low and I'm sure the go-arounds are loud if you happen to be in the right(/wrong) place. But I'd advise people to take the suggestion that this causes any meaningful disruption to "a very busy area of the city" with a pinch of salt.

Having lived in that area for a while and worked for airbus, yes, that, and additionally landing in finkenwerder isn't the easiest sometimes, the relatively (to other airpors in the area) approach and the changing water/city body together with the beluga not being the most stable aircraft either. Some of their landing look like crosswind hurricane while standing at the airfield you wonder if there is any wind going at all...

@ChrisH Othmarschen and Altona in general is a very busy area of the city. It has high population density and a lot of schools. The landings themselves already cause disruption (~4 or 6 landings a day low over the area), the go arounds are even worse for it

@PlasmaHH I was wondering also if it had something to do with container ships? I imagine that a Beluga cannot land if there is a big container ship passing on the Elbe

@JoeBentley occasionally a go-around will be caused by a container ship, but I can't imagine it will happen frequently. Those ships move in a predictable way and are tracked by AIS, so the tower can plan accordingly.

Many parts of "Altona in general" are indeed busy. Altona-Altstadt, Altona-Nord, Ottensen, those are busy areas. Upwards of 10k people per square km, lots of shops, cafes, restaurants, bars, theatres, etc. Othmarschen has a quarter of that population density, far fewer shops, practically no entertainment venues. I know the area well, I lived there for 6 or 7 years. If that part of town is "very busy" I dread to think how many intensifiers you'd need for central Altona, nevermind for the city centre.

@JoeBentley interesting question. I think that with maybe 60-70m cruise ships would even be higher. However I don't think the tower will clear them to land if there is any big ship near

@ChrisH The landings also disrupt Ottensen

I speak from many years experience living in Ottensen: they absolutely do not.


text:
Why do early airliners have a "backward slope" windshield?
Does anyone know why many 1930's airliners in the US have cockpit windows that slope down and rearward, the opposite of what you might expect?
Consider the Boeing 247:

Source
Or the Lockheed Electra Model 10:

enter link description here
Or the Ford Trimotor:

Source
This seems to have disappeared shortly after, the DC-2 and on, for instance, lack this feature.

It's going to be either rain or sun reflection related.

I´m pretty sure the Boeing 247 already appeared in an answer by Peter Kämpf at some point, and it was indeed due to someone´s pet obsession with anti-glare features for the cockpit, at the expense of all reason. Can´t find the exact post now, sadly.

@AEhere this one? https://aviation.stackexchange.com/a/25104/609

@ratchetfreak yes, that one!

Most likely because of the application of boat design patterns to the design of aircraft of the day.

@ratchetfreak - yes, that does seem to be it! Sorry, it's aehere's post.

Although the 247 was mentioned in another answer as noted above, it was an answer to a very different question and I think this question deserves its own answer even if the answer copies material from the other answer or directs readers to the other answer or mentions that the content originally appeared in another answer.

If it's a good idea here, why not in the cockpit of an aeroplane, right?
Image source
The forward slope was designed to avoid reflection of lighted instruments in the window panes, but that turned out to create other reflections. From this site:

The cockpit windshield of the first 247s was angled "forward" instead of the conventional aft sweep. This was the design solution (similar to that adopted by other contemporary aircraft that used a forward raked windscreen) to the problem of lighted control panel instruments reflecting off the windshield at night, but it turned out that the forward-sloping windshield would reflect ground lights instead, especially during landings, and it also increased drag slightly.[12][13] By the introduction of the 247D, the windshield was sloped aft in the usual way, and the night-glare problem was resolved by installing an extension (the glarescreen) over the control panel.[14]

Also mentioned in this answer.

I'm trying to figure out how a forward-sloping windshield could possibly reflect ground lights in a manner that the pilots would care about. My best guess is that they meant it would reflect light from the ceiling, illuminated by ground lights.

@TannerSwett Yeah it could very well be the reason stated in Peter Kämpf's answer, limitation of upwards visibility.


text:
Does a 3-blade prop handle up/down drafts better than a 2-blade?
I'm about to order a plane kit (RV-10) and one of the choices I'll be making is a 3-blade vs 2-blade propeller. A 3-blade prop should be: easier to balance, quieter and capable of generating more thrust - assuming engine has adequate power - which helps to shorten the take-off distance. However, it also adds more drag (i.e. it's less efficient) so it's common to see a reduction of top-speed.
Can anyone comment on their relative performance in winds or high density altitudes? I'm not concerned about the top-speed as much as dealing with adverse weather. I plan to fly in or near the Rockies. If benefits are negligible, the cost savings are considerable when going with a 2-blade prop. I reached out to Hartzell but have not heard back.

I think the answer can be found in this question and in this one

The benefit of the 3 blade is noise level and ground clearance. If those are critical features, spend the money.  For pure efficiency, a 2 blade will give more static thrust.  In the bush, Beaver and C-185 operators like 3 blade props because it eliminates the searing BLAAAT sound on takeoff from supersonic tips and reduces complaints by the neighbours.  If you can afford to build an RV-10 in the first place, the cost difference probably isn't a show stopper.

Seems like rate of climb and takeoff should be priorities (high DAs?).  I'd go with a three blade.  A few more knots in cruise out of a two blade really doesn't do much fore me.


text:
What happens to the spinning wheels of the landing gear after takeoff, before retraction?
What happens to the spinning landing gear right after V2 on takeoff? Are they automatically or manually braked to a halt before gear retraction? Or are they allowed to spin to a halt in the wheel wells?
I am asking this question in the context of common passenger planes, such as the Boeing 737-800 or the Airbus A320.

Useful information here: Will landing gears with wheel fairings keep spinning due to airflow?

The question was formulated in the context of a jet liner such as the Boeing 737-800 or the Airbus, I do not understand why there are persons responding to this question in the context of a smaller less complex aircraft.
With that being said, I will like to expand the answer by DeltaLima. Boeing uses a device attached to the Brake Metering Valve Module called a De-spin Actuator. This device directs controlled hydraulic fluid to the brakes when the landing gear lever is set to retract to stop the spinning of the wheels. More information can be found in the AMM ATA 32 (Aircraft Maintenance Manual Chapter on Landing Gear).
https://www.facebook.com/aviationisawesome/videos/1598830786833154/
Video of a 737-500 from inside the main gear wheel well shows:

After becoming airborne, the landing gear suspension unloads and the wheel hangs down, still spinning
Once the main gear arms begin retracting, the wheel stops within a second
The gear has fulled retracted in 6-7 seconds.

De-Spin Actuator
Source: http://www.eaton.com/ecm/groups/public/@pub/@eaton/@aero/documents/content/ct_194202.pdf

Welcome to Aviation.SE Andres! Very interesting answer, I hope you come back with more of those.

I especially like how you can see the RH main wheels "push" downwards when the aircraft rotates.  Very nice video; thanks for sharing (and answering).

Video is unavailable. In the future, it would be nice if the content of the video are described so that if a link goes bad, the answer here is still useful.

@mahboudz agreed, I've managed to find a copy at https://www.facebook.com/aviationisawesome/videos/1598830786833154/UzpfSTIzNjk3MDc0NjY5OjEwMTU2MzU4MDQ0ODE5Njcw/, and I submitted an edit to this post to add a description of what happens in the video.

In most common passenger aircraft brakes are automatically applied when the gear is retracted. The wheels stop spinning before they enter the wheel well. This is particularly important because a spinning damaged tire may cause damage to hydraulic and fuel lines which are usually routed near the wheel bay.
The nose wheels have no brakes, so they are spinning while they are being retracted. Most aircraft have rubber brake pads (snubbers) fitted in the nose gear bay that stop the spinning of the gear once the wheels are retracted.

These are the "nose wheel spin brake linings" (the braking pads) that stop the spinning of the nose wheel after retraction.

General policy when operating a retractable gear aircraft is to apply brake pressure after becoming airborne and prior to commanding the retraction of the landing gear.  This causes the wheel to cease spinning, removing gyroscopic precession loads on the landing gear during retraction (remember a spinning tire acts like a big gyroscope and wants to maintain its orientation in space) as well as reduces the chances of damaging components inside the wheel well when the gear is stowed.

What does this answer add that isn't in any of the other answers?

Correctly noting that spinning wheels create gyroscopic precession loads on the gear during retraction?

Which is already noted in another answer https://aviation.stackexchange.com/a/42245/15982

No they didn't.  That poster described it as centrifugal force.  It is not.

Could you have notified the poster of the misnomer in a comment to their answer?

No.  I make my own comments in this case.

I fly retracts and have heard this question a dozen times. Some people say to tap the brakes, other people (like me) never do. I have asked certified mechanics, and they have all said it doesn't matter.
My advice is to follow the POH. If it says tap the brakes, then do so. Otherwise do or don't it doesn't matter.

I was checked out in a Cessna 172RG Cutlass, and the POH did indeed specify applying brakes before retracting gear.

There you go. The 177 I flew probably said the same thing but I don't recall.

On the 737, the main gear gets gear retract braking pressure when the gear lever is placed in the UP position. It also doesn't hurt that a heavy brake stack will close up under gravity as the wheels are turned horizontal during retract. The nose wheels are free to spin up into the wheel well, where they contact a friction brake that drags on the tires; slowing and stopping nose wheel spin.

Follow the POH is always the best and most correct answer because it is different on different models and types.
I have a 1971 C177RG POH in front of me right now and it says:
Landing Gear Retraction
Before retracting the landing gear, the brakes should be applied momentarily to stop wheel rotation

Yabut in FAA parlance "should be" is not required, and "must be" or "shall be" is required.  So this POH language is a suggestion not a requirement.

Actually, the main reason brakes are applied to stop spinning wheels is the centripetal and centrifugal forces. These forces are a reaction to the spinning of wheels. As the gear is retracted, the forces that would be generated would cause uncontrollable vibrations throughout the aircraft. The result would be catastrophic. Want an example of how this works, find a fidget spinner. While it is spinning, rotate the spinner on an axis. You will feel the spinner "wobble". Now imagine the landing gear and the speed they are rotating at.

I believe you mean gyroscopic forces.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Is the Boeing AC-25A Banshee a real concept or a gimmick?
Could this be a feasible concept aircraft to replace the AC-130?

I think "gimmick" would be a step up for this bit of nonsense, honestly.

Well, the URL should give it away already ...
The AC-130 is a turboprop aircraft with relatively low wing loading. It can be operated from (almost) unprepared airstrips.
Now compare it to the 747-400: Higher minimum speed means more distance to the taget, higher tire pressure means operation from established airbases with a long runway. Just look at the T-tail: Such a change would necessitate a complete redesign of the rear fuselage and the vertical tail. With the changes photoshopped into the picture the "AC-25" would be 80% new and could use only a few parts from the 747.
Here is an incomplete list of what needs to be changed:

Lifting the wing up to a high wing position would need less dihedral (or even some anhedral, to be precise), so the wing needs major modifications
The landing gear would need a complete redesign. Currently, some parts retract into the wing, which is no longer there. The pods at the side of the fuselage are too small to hold the current gear.
The mid fuselage section is a new design, too. Moving the wing up means a complete redesign of the bulkheads and wing attachment points. Also, fuel lines and cables need to be re-routed.

URL was the first thing I looked at (I tend to look at URLs before clicking - it's a habit).  Someone's made a cute JPG and pasted a Boeing logo on it.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
How to find Designated Pilot Examiners (DPE)s?
I took my first private pilot exam three weeks ago, and schedule a re-take with the  Designated Pilot Examiner (DPE) for next week. Except this week, he called and cancelled. :(
I've called and texted every DPE in the state of Utah, trying to make any appointment within the FAA's 60-day window for a retake. (https://designee.faa.gov/ though there's a fair number listed that only do other stuff, like helicopters)
No responses. What should I do?
Do I look at other states? My instructor doesn't want me to have my exam at an unfamiliar airport, but I don't see what the other options are? Do I figure out alternative ways of contact (home addresses?) Do I talk to the FSDO? Is there someone "in charge"?
Is there someone that can help me? I feel like I'm doing something wrong.
EDIT: Rescheduled with another DPE, and then he canceled day-of. :/

@RTO Unfortunately my instructor is on vacation, and only somewhat available. Pilots seems to have ... erratic ... schedules.

@RTO thanks I appreciate it.

What state are you in?


text:
What could be the purpose of multiple mount positions for the rear fins of a (turbine-powered) missile?
The official photos of the Iranian 358 missile have an interesting detail. Apparently the rear fins can be mounted at different positions on the body, given the multiple mounts.

I have circled the extra mounts visible. What (flexibility?) could be gained by doing so, i.e. varying the mount position of the rear fins?
(The 358 is powered by a gas turbine, making it a "loitering SAM" against drones, according to Wikipedia, although I don't know if that has any relationship with my Q.)

Other images of the 358 show multiple fins attached. Not just the normal 2 sets, 1 front, 1 rear.
(I too am having issues posting a pic, but this link)
https://militarymatters.online/defense-news/358-vs-scan-eagle-anti-drone-action/
I surmise using the center set of fins requires moving the rear ones to the very back position.
Why do they need all those? Unknown, but possibly different flight profiles need different surfaces.

Actually, the last image in the latter article shows a single set of rear mounts, and also a different shape of the rear fins. So, possibly the official photos were showing some prototype and they'd not yet decided on the rear fins shape or their exact position.

The photo in that article is from a specimen captured in 2020 on dhow see p.8 https://www.dia.mil/Portals/110/Documents/News/Military_Power_Publications/Seized_at_Sea.pdf OTOH from the same source, the photo in my Q is from a 2023 expo. So, another explanation is that they might have introduced the extra mount points for the rear fins, as well as alternate shape fins later on. The expo fins are far less angled [compared to the 2020 ones], so this might give the missile more flight time, due to extra lift.